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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

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Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A021009 Triangle of coefficients of Laguerre polynomials n!*L_n(x) (rising powers of x).

Original entry on oeis.org

1, 1, -1, 2, -4, 1, 6, -18, 9, -1, 24, -96, 72, -16, 1, 120, -600, 600, -200, 25, -1, 720, -4320, 5400, -2400, 450, -36, 1, 5040, -35280, 52920, -29400, 7350, -882, 49, -1, 40320, -322560, 564480, -376320, 117600, -18816, 1568, -64, 1, 362880, -3265920
Offset: 0

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Author

Keywords

Comments

In absolute values, this sequence also gives the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals (j-1)^2 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
A partial permutation on a set X is a bijection between two subsets of X. |T(n,n-k)| equals the numbers of partial permutations of an n-set having domain cardinality equal to k. Let E denote the operator D*x*D, where D is the derivative operator d/dx. Then E^n = Sum_{k = 0..n} |T(n,k)|*x^k*D^(n+k). - Peter Bala, Oct 28 2008
The unsigned triangle is the generalized Riordan array (exp(x), x) with respect to the sequence n!^2 as defined by Wang and Wang (the generalized Riordan array (exp(x), x) with respect to the sequence n! is Pascal's triangle A007318, and with respect to the sequence n!*(n+1)! is A105278). - Peter Bala, Aug 15 2013
The unsigned triangle appears on page 83 of Ser (1933). - N. J. A. Sloane, Jan 16 2020

Examples

			The triangle a(n,m) starts:
n\m   0       1      2       3      4      5    6  7  8
0:    1
1:    1      -1
2:    2      -4      1
3:    6     -18      9      -1
4:   24     -96     72     -16      1
5:  120    -600    600    -200     25     -1
6:  720   -4320   5400   -2400    450    -36    1
7: 5040  -35280  52920  -29400   7350   -882   49  -1
8:40320 -322560 564480 -376320 117600 -18816 1568 -64 1
...
From _Wolfdieter Lang_, Jan 31 2013 (Start)
Recurrence (usual one): a(4,1) = 7*(-18) - 6 - 3^2*(-4) = -96.
Recurrence (simplified version): a(4,1) = 5*(-18) - 6 = -96.
Recurrence (Sage program): |a(4,1)| = 6 + 3*18 + 4*9 = 96. (End)
Embedded recurrence (Maple program): a(4,1) = -4!*(1 + 3) = -96.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 799.
  • G. Rota, Finite Operator Calculus, Academic Press, New York, 1975.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 83.

Crossrefs

Row sums give A009940, alternating row sums are A002720.
Column sequences (unsigned): A000142, A001563, A001809-A001812 for m=0..5.
Central terms: A295383.
For generators and generalizations see A132440.

Programs

  • Magma
    /* As triangle: */ [[((-1)^k)*Factorial(n)*Binomial(n, k)/Factorial(k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 18 2020
  • Maple
    A021009 := proc(n,k) local S; S := proc(n,k) option remember; `if`(k = 0, 1, `if`( k > n, 0, S(n-1,k-1)/k + S(n-1,k))) end: (-1)^k*n!*S(n,k) end: seq(seq(A021009(n,k), k=0..n), n=0..8); # Peter Luschny, Jun 21 2017
    # Alternative for the unsigned case (function RiordanSquare defined in A321620):
    RiordanSquare(add(x^m, m=0..10), 10, true); # Peter Luschny, Dec 06 2018
  • Mathematica
    Flatten[ Table[ CoefficientList[ n!*LaguerreL[n, x], x], {n, 0, 9}]] (* Jean-François Alcover, Dec 13 2011 *)
  • PARI
    p(n) = denominator(bestapprPade(Ser(vector(2*n, k, (k-1)!))));
    concat(1, concat(vector(9, n, Vec(-p(n)))))  \\ Gheorghe Coserea, Dec 01 2016
    
  • PARI
    {T(n, k) = if( n<0, 0, n! * polcoeff( sum(i=0, n, binomial(n, n-i) * (-x)^i / i!), k))}; /* Michael Somos, Dec 01 2016 */
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n)); \\ Michel Marcus, Feb 06 2021
    
  • Sage
    def A021009_triangle(dim): # computes unsigned T(n,k).
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+(2*k+1)*M[n-1,k]+(k+1)^2*M[n-1,k+1]
        return M
    A021009_triangle(9) # Peter Luschny, Sep 19 2012
    

Formula

a(n, m) = ((-1)^m)*n!*binomial(n, m)/m! = ((-1)^m)*((n!/m!)^2)/(n-m)! if n >= m, otherwise 0.
E.g.f. for m-th column: (-x/(1-x))^m /((1-x)*m!), m >= 0.
Representation (of unsigned a(n, m)) as special values of Gauss hypergeometric function 2F1, in Maple notation: n!*(-1)^m*hypergeom([ -m, n+1 ], [ 1 ], 1)/m!. - Karol A. Penson, Oct 02 2003
Sum_{m>=0} (-1)^m*a(n, m) = A002720(n). - Philippe Deléham, Mar 10 2004
E.g.f.: (1/(1-x))*exp(x*y/(x-1)). - Vladeta Jovovic, Apr 07 2005
Sum_{n>=0, m>=0} a(n, m)*(x^n/n!^2)*y^m = exp(x)*BesselJ(0, 2*sqrt(x*y)). - Vladeta Jovovic, Apr 07 2005
Matrix square yields the identity matrix: L^2 = I. - Paul D. Hanna, Nov 22 2008
From Tom Copeland, Oct 20 2012: (Start)
Symbolically, with D=d/dx and LN(n,x)=n!L_n(x), define :Dx:^j = D^j x^j, :xD:^j = x^j D^j, and LN(.,x)^j = LN(j,x) = row polynomials of A021009.
Then some useful relations are
1) (:Dx:)^n = LN(n,-:xD:) [Rodriguez formula]
2) (xDx)^n = x^n D^n x^n = x^n LN(n,-:xD:) [See Al-Salam ref./A132440]
3) (DxD)^n = D^n x^n D^n = LN(n,-:xD:) D^n [See ref. in A132440]
4) umbral composition LN(n,LN(.,x))= x^n [See Rota ref.]
5) umbral comp. LN(n,-:Dx:) = LN(n,-LN(.,-:xD:)) = 2^n LN(n,-:xD:/2)= n! * (n-th row e.g.f.(x) of A038207 with x replaced by :xD:).
An example for 2) is the operator (xDx)^2 = (xDx)(xDx) = xD(x^2 + x^3D)= 2x^2 + 4x^3 D + x^4 D^2 = x^2 (2 + 4x D + x^2 D^2) = x^2 (2 + 4 :xD: + :xD:^2) = x^2 LN(2,-:xD:) = x^2 2! L_2(-:xD:).
An example of the umbral composition in 5) is given in A038207.
The op. xDx is related to the Euler/binomial transformation for power series/o.g.f.s. through exp(t*xDx) f(x) = f[x/(1-t*x)]/(1-t*x) and to the special Moebius/linear fractional/projective transformation z exp(-t*zDz)(1/z)f(z) = f(z/(1+t*z)).
For a general discussion of umbral calculus see the Gessel link. (End)
From Wolfdieter Lang, Jan 31 2013: (Start)
Standard recurrence derived from the three term recurrence of the orthogonal polynomials system {n!*L(n,x)}: L(n,x) = (2*n - 1 - x)*L(n-1,x) - (n-1)^2*L(n-2,x), n>=1, L(-1,x) = 0, L(0,x) = 1.
a(n,m) = (2*n-1)*a(n-1,m) - a(n-1,m-1) - (n-1)^2*a(n-2,m),
n >=1, with a(n,-1) = 0, a(0,0) = 1, a(n,m) = 0 if n < m. (compare this with Peter Luschny's program for the unsigned case |a(n,m)| = (-1)^m*a(n,m)).
Simplified recurrence (using column recurrence from explicit form for a(n,m) given above):
a(n,m) = (n+m)*a(n-1,m) - a(n-1,m-1), n >= 1, a(0,0) = 1, a(n,-1) = 0, a(n,m) = 0 if n < m. (End)
|T(n,k)| = [x^k] (-1)^n*U(-n,1,-x), where U(a,b,x) is Kummer's hypergeometric U function. - Peter Luschny, Apr 11 2015
T(n,k) = (-1)^k*n!*S(n,k) where S(n,k) is recursively defined by: "if k = 0 then 1 else if k > n then 0 else S(n-1,k-1)/k + S(n-1,k)". - Peter Luschny, Jun 21 2017
The unsigned case is the exponential Riordan square (see A321620) of the factorial numbers. - Peter Luschny, Dec 06 2018
Omitting the diagonal and signs, this array is generated by the commutator [D^n,x^n] = D^n x^n - x^n D^n = Sum_{i=0..n-1} ((n!/i!)^2/(n-i)!) x^i D^i on p. 9 of both papers by Belov-Kanel and Kontsevich. - Tom Copeland, Jan 23 2020

Extensions

Name changed and table given by Wolfdieter Lang, Nov 28 2011

A028421 Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 6, 22, 18, 4, 24, 100, 105, 40, 5, 120, 548, 675, 340, 75, 6, 720, 3528, 4872, 2940, 875, 126, 7, 5040, 26136, 39396, 27076, 9800, 1932, 196, 8, 40320, 219168, 354372, 269136, 112245, 27216, 3822, 288, 9
Offset: 0

Views

Author

Peter Wiggen (wiggen(AT)math.psu.edu)

Keywords

Comments

Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.
We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2 + 6*n + 3*n^2)/x^2 - (6 + 22*n + 18*n^2 + 4*n^3)/x^3 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.
The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).
(End)
For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - Tom Copeland, Feb 28 2014
From Wolfdieter Lang, Nov 25 2018: (Start)
The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).
The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).
The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)
The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
n\k       0        1        2        3        4       5       6      7     8   9 10
------------------------------------------------------------------------------------
0:        1
1:        1        2
2:        2        6        3
3:        6       22       18        4
4:       24      100      105       40        5
5:      120      548      675      340       75       6
6:      720     3528     4872     2940      875     126       7
7:     5040    26136    39396    27076     9800    1932     196      8
8:    40320   219168   354372   269136   112245   27216    3822    288     9
9:   362880  2053152  3518100  2894720  1346625  379638   66150   6960   405  10
10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11
... - _Wolfdieter Lang_, Nov 23 2018
		

Crossrefs

Row sums give A000254(n+1), n >= 0.
Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.
From Johannes W. Meijer, Oct 07 2009: (Start)
A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.
The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).
Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

Programs

  • Maple
    A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:
    seq(seq(A028421(n,k), k=0..n), n=0..8);
    # Johannes W. Meijer, Oct 07 2009, Revised Sep 09 2012
    egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):
    ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):
    seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # Peter Luschny, Jun 12 2022
  • Mathematica
    f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(-ln(1 - x), 10, True) # Peter Luschny, Jan 03 2019

Formula

E.g.f.: d/dt(-log(1-t)/(1-t)^x). - Vladeta Jovovic, Oct 12 2003
The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - Peter Bala, Jul 18 2013
Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - Wolfdieter Lang, Nov 25 2018

Extensions

Edited by Wolfdieter Lang, Nov 23 2018

A147703 Triangle [1,1,1,0,0,0,...] DELTA [1,0,0,0,...] with Deléham DELTA defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 13, 27, 20, 7, 1, 34, 80, 73, 35, 9, 1, 89, 234, 252, 151, 54, 11, 1, 233, 677, 837, 597, 269, 77, 13, 1, 610, 1941, 2702, 2225, 1199, 435, 104, 15, 1, 1597, 5523, 8533, 7943, 4956, 2158, 657, 135, 17, 1
Offset: 0

Views

Author

Paul Barry, Nov 10 2008

Keywords

Comments

Equal to A062110*A007318 when A062110 is regarded as a triangle read by rows.

Examples

			Triangle begins
   1;
   1,   1;
   2,   3,   1;
   5,   9,   5,   1;
  13,  27,  20,   7,  1;
  34,  80,  73,  35,  9,  1;
  89, 234, 252, 151, 54, 11, 1;
		

Crossrefs

Row sums are A006012. Diagonal sums are A147704.

Programs

  • Maple
    # The function RiordanSquare is defined in A321620:
    RiordanSquare(1 / (1 - x / (1 - x / (1 - x))), 10); # Peter Luschny, Jan 26 2020
  • Mathematica
    nmax=9; Flatten[CoefficientList[Series[CoefficientList[Series[(1 - 2*x)/(1 - (3 + y)*x + (1 + y)*x^2), {x, 0, nmax}], x], {y, 0, nmax}], y]] (* Indranil Ghosh, Mar 11 2017 *)

Formula

Riordan array ((1-2x)/(1-3x+x^2), x(1-x)/(1-3x+x^2)).
Sum_{k=0..n} T(n,k)*x^k = A152239(n), A152223(n), A152185(n), A152174(n), A152167(n), A152166(n), A152163(n), A000007(n), A001519(n), A006012(n), A081704(n), A082761(n), A147837(n), A147838(n), A147839(n), A147840(n), A147841(n), for x = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Dec 01 2008
G.f.: (1-2*x)/(1-(3+y)*x+(1+y)*x^2). - Philippe Deléham, Nov 26 2011
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), for n > 1. - Philippe Deléham, Feb 12 2012
The Riordan square of the odd indexed Fibonacci numbers A001519. - Peter Luschny, Jan 26 2020

A063967 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k) + T(n-1,k-1) + T(n-2,k-1) and T(0,0) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 7, 5, 1, 5, 15, 16, 7, 1, 8, 30, 43, 29, 9, 1, 13, 58, 104, 95, 46, 11, 1, 21, 109, 235, 271, 179, 67, 13, 1, 34, 201, 506, 705, 591, 303, 92, 15, 1, 55, 365, 1051, 1717, 1746, 1140, 475, 121, 17, 1, 89, 655, 2123, 3979, 4759, 3780, 2010, 703, 154, 19, 1
Offset: 0

Views

Author

Henry Bottomley, Sep 05 2001

Keywords

Examples

			T(3,1) = T(2,1) + T(1,1) + T(2,0) + T(1,0) = 3 + 1 + 2 + 1 = 7.
Triangle begins:
   1,
   1,   1,
   2,   3,   1,
   3,   7,   5,   1,
   5,  15,  16,   7,   1,
   8,  30,  43,  29,   9,   1,
  13,  58, 104,  95,  46,  11,  1,
  21, 109, 235, 271, 179,  67, 13,  1,
  34, 201, 506, 705, 591, 303, 92, 15, 1
		

Crossrefs

Row sums are A002605.
Columns include: A000045(n+1), A023610(n-1).
Main diagonal: A000012, a(n, n-1) = A005408(n-1).
Matrix inverse: A091698, matrix square: A091700.
Cf. A321620.
Sum_{k=0..n} x^k*T(n,k) is (-1)^n*A057086(n) (x=-11), (-1)^n*A057085(n+1) (x=-10), (-1)^n*A057084(n) (x=-9), (-1)^n*A030240(n) (x=-8), (-1)^n*A030192(n) (x=-7), (-1)^n*A030191(n) (x=-6), (-1)^n*A001787(n+1) (x=-5), A000748(n) (x=-4), A108520(n) (x=-3), A049347(n) (x=-2), A000007(n) (x=-1), A000045(n) (x=0), A002605(n) (x=1), A030195(n+1) (x=2), A057087(n) (x=3), A057088(n) (x=4), A057089(n) (x=5), A057090(n) (x=6), A057091(n) (x=7), A057092(n) (x=8), A057093(n) (x=9). - Philippe Deléham, Nov 03 2006

Programs

  • Haskell
    a063967_tabl = [1] : [1,1] : f [1] [1,1] where
       f us vs = ws : f vs ws where
         ws = zipWith (+) ([0] ++ us ++ [0]) $
              zipWith (+) (us ++ [0,0]) $ zipWith (+) ([0] ++ vs) (vs ++ [0])
    -- Reinhard Zumkeller, Apr 17 2013
  • Mathematica
    T[n_, k_] := Sum[Binomial[j, n - j]*Binomial[j, k], {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 11 2017, after Paul Barry *)
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[1/(1 - x - x^2), 11] // Flatten (* Peter Luschny, Nov 27 2018 *)

Formula

G.f.: 1/(1-x*(1+x)*(1+y)). - Vladeta Jovovic, Oct 11 2003
Riordan array (1/(1-x-x^2), x(1+x)/(1-x-x^2)). The inverse of the signed version (1/(1+x-x^2),x(1-x)/(1+x-x^2)) is abs(A091698). - Paul Barry, Jun 10 2005
T(n, k) = Sum_{j=0..n} C(j, n-j)C(j, k). - Paul Barry, Nov 09 2005
Diagonal sums are A002478. - Paul Barry, Nov 09 2005
A026729*A007318 as infinite lower triangular matrices. - Philippe Deléham, Dec 11 2008
Central coefficients T(2*n,n) are A137644. - Paul Barry, Apr 15 2010
Product of Riordan arrays (1, x(1+x))*(1/(1-x), x/(1-x)), that is, A026729*A007318. - Paul Barry, Mar 14 2011
Triangle T(n,k), read by rows, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 12 2011

A155161 A Fibonacci convolution triangle: Riordan array (1, x/(1 - x - x^2)). Triangle T(n,k), 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 3, 5, 3, 1, 0, 5, 10, 9, 4, 1, 0, 8, 20, 22, 14, 5, 1, 0, 13, 38, 51, 40, 20, 6, 1, 0, 21, 71, 111, 105, 65, 27, 7, 1, 0, 34, 130, 233, 256, 190, 98, 35, 8, 1, 0, 55, 235, 474, 594, 511, 315, 140, 44, 9, 1, 0, 89, 420, 942, 1324, 1295, 924, 490, 192, 54, 10, 1
Offset: 0

Views

Author

Philippe Deléham, Jan 21 2009

Keywords

Examples

			Triangle begins:
[0] 1;
[1] 0,  1;
[2] 0,  1,   1;
[3] 0,  2,   2,   1;
[4] 0,  3,   5,   3,   1;
[5] 0,  5,  10,   9,   4,   1;
[6] 0,  8,  20,  22,  14,   5,  1;
[7] 0, 13,  38,  51,  40,  20,  6,  1;
[8] 0, 21,  71, 111, 105,  65, 27,  7, 1;
[9] 0, 34, 130, 233, 256, 190, 98, 35, 8, 1.
		

Crossrefs

Row sums are in A215928.
Central terms: T(2*n,n) = A213684(n) for n > 0.

Programs

  • Haskell
    a155161 n k = a155161_tabl !! n !! k
    a155161_row n = a155161_tabl !! n
    a155161_tabl = [1] : [0,1] : f [0] [0,1] where
       f us vs = ws : f vs ws where
         ws = zipWith (+) (us ++ [0,0]) $ zipWith (+) ([0] ++ vs) (vs ++ [0])
    -- Reinhard Zumkeller, Apr 17 2013
  • Maple
    T := (n, k) -> binomial(n-1, k-1)*hypergeom([-(n-k)/2, -(n-k-1)/2], [1-n], -4):
    seq(seq(simplify(T(n, k)), k = 0..n), n = 0..11); # Peter Luschny, May 23 2021
    # Uses function PMatrix from A357368.
    PMatrix(10, n -> combinat:-fibonacci(n)); # Peter Luschny, Oct 07 2022
  • Mathematica
    CoefficientList[#, y]& /@ CoefficientList[(1-x-x^2)/(1-x-x^2-x*y)+O[x]^12, x] // Flatten (* Jean-François Alcover, Mar 01 2019 *)
    (* Generates the triangle without the leading '1' (rows are rearranged). *)
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[x/(1 - x - x^2), 11] // Flatten  (* Peter Luschny, Feb 27 2021 *)
  • Maxima
    M(n,k):=pochhammer(n,k)/k!;
    create_list(sum(M(k,i)*binomial(i,n-i-k),i,0,n-k),n,0,8,k,0,n); /* Emanuele Munarini, Mar 15 2011 */
    

Formula

T(n, k) given by [0,1,1,-1,0,0,0,...] DELTA [1,0,0,0,...] where DELTA is the operator defined in A084938.
a(n,k) = Sum_{i=0..n-k} M(k,i)*binomial(i,n-i-k), where M(n,k) = n(n+1)(n+2)...(n+k-1)/k!. - Emanuele Munarini, Mar 15 2011
Recurrence: a(n+2,k+1) = a(n+1,k+1) + a(n+1,k) + a(n,k+1). - Emanuele Munarini, Mar 15 2011
G.f.: (1-x-x^2)/(1-x-x^2-x*y). - Philippe Deléham, Feb 08 2012
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000129(n) (n > 0), A052991(n), A155179(n), A155181(n), A155195(n), A155196(n), A155197(n), A155198(n), A155199(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Feb 08 2012
T(n, k) = binomial(n-1, k-1)*hypergeom([-(n-k)/2, -(n-k-1)/2], [1-n], -4). - Peter Luschny, May 23 2021

A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1
Offset: 0

Views

Author

Paul D. Hanna, May 02 2006

Keywords

Comments

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

Examples

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.
		

Crossrefs

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Programs

  • Mathematica
    (* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
    z = 8; a = 1; b = 1; c = 1; d = 1;
    p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
    t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
    w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
    g[n_] := CoefficientList[w[n, -x], x];
    TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
    Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
    TableForm[Table[g[n], {n, -1, z}]]
    Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
    (* Clark Kimberling, Aug 04 2011 *)
    T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)
  • PARI
    {T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* Chebyshev Polynomials as Antidiagonals: */
    {T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • Sage
    # uses[riordan_square from A321620]
    # Computes the unsigned triangle.
    riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

Formula

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + .... Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1
Offset: 0

Views

Author

Philippe Deléham, Jun 09 2007

Keywords

Comments

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

Examples

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020
		

Crossrefs

Programs

  • Maple
    # The function RiordanSquare is defined in A321620.
    RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
    LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019
  • Mathematica
    max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)
  • Sage
    @CachedFunction
    def A129818(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
        return A129818(n-1,k-1) - A129818(n-2,k) - h
    for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

A322942 Jacobsthal triangle, coefficients of orthogonal polynomials J(n, x) where J(n, 0) are the Jacobsthal numbers (A001045 with a(0) = 1). Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 5, 3, 1, 5, 12, 10, 4, 1, 11, 27, 28, 16, 5, 1, 21, 62, 75, 52, 23, 6, 1, 43, 137, 193, 159, 85, 31, 7, 1, 85, 304, 480, 456, 290, 128, 40, 8, 1, 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1, 341, 1442, 2793, 3336, 2750, 1652, 742, 248, 61, 10, 1
Offset: 0

Views

Author

Peter Luschny, Jan 03 2019

Keywords

Comments

The name 'Jacobsthal triangle' used here is not standard.

Examples

			The first few polynomials are:
  J(0, x) = 1;
  J(1, x) = x + 1;
  J(2, x) = x^2 + 2*x + 1;
  J(3, x) = x^3 + 3*x^2 +  5*x + 3;
  J(4, x) = x^4 + 4*x^3 + 10*x^2 + 12*x + 5;
  J(5, x) = x^5 + 5*x^4 + 16*x^3 + 28*x^2 + 27*x + 11;
  J(6, x) = x^6 + 6*x^5 + 23*x^4 + 52*x^3 + 75*x^2 + 62*x + 21;
The triangle starts:
  [0]   1;
  [1]   1,   1;
  [2]   1,   2,    1;
  [3]   3,   5,    3,    1;
  [4]   5,  12,   10,    4,   1;
  [5]  11,  27,   28,   16,   5,   1;
  [6]  21,  62,   75,   52,  23,   6,   1;
  [7]  43, 137,  193,  159,  85,  31,   7,  1;
  [8]  85, 304,  480,  456, 290, 128,  40,  8, 1;
  [9] 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1;
		

Crossrefs

Row sums are A152035, alternating row sums are A000007, values at x=1/2 are A323232, values at x=0 (first column) are A152046.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
    g:= func< n,x | (&+[Binomial(n-k,k)*2^k*(x+1)^(n-2*k): k in [0..Floor(n/2)]]) >;
    f:= func< n,x | n le 1 select (x+1)^n else g(n,x) - 2*g(n-2,x) >;
    A322942:= func< n,k | Coefficient(R!( f(n,x) ), k) >;
    [A322942(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 20 2023
  • Maple
    J := proc(n) option remember;
    `if`(n < 3, [1, x+1, x^2 + 2*x + 1][n+1], (x+1)*J(n-1) + 2*J(n-2));
    sort(expand(%)) end: seq(print(J(n)), n=0..11); # Computes the polynomials.
    seq(seq(coeff(J(n), x, k), k=0..n), n=0..11);
  • Mathematica
    J[n_, x_]:= J[n, x]= If[n<3, (x+1)^n, (x+1)*J[n-1, x] + 2*J[n-2, x]];
    T[n_, k_]:= Coefficient[J[n, x], x, k];
    Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 17 2019 *)
    (* Second program *)
    A322942[n_, k_]:= Coefficient[Series[Boole[n==0] + (I*Sqrt[2])^n*(ChebyshevU[n, (x+1)/(2*Sqrt[2]*I)] + ChebyshevU[n-2, (x+ 1)/(2*Sqrt[2]*I)]), {x,0,50}], x, k];
    Table[A322942[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 20 2023 *)
  • Sage
    # use[riordan_square from A321620]
    riordan_square((2*x^2 - 1)/((x + 1)*(2*x - 1)), 9)
    

Formula

J(n, x) = (x+1)*J(n-1, x) + 2*J(n-2, x) for n >= 3.
T(n, k) = [x^k] J(n, x).
Equals the Riordan square (cf. A321620) generated by (2*x^2-1)/((x + 1)*(2*x - 1)).
Sum_{k=0..n} T(n, k) = A152035(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
From G. C. Greubel, Sep 20 2023: (Start)
T(n, k) = [x^k]( [n=0] + (i*sqrt(2))^n*(ChebyshevU(n, (x+1)/(2*sqrt(2)*i)) + ChebyshevU(n-2, (x+1)/(2*sqrt(2)*i))) ).
G.f.: (1 - 2*t^2)/(1 - (x+1)*t - 2*t^2).
Sum_{k=0..floor(n/2)} T(n-k, k) = (2/3)*[n=0] + A006138(n-1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = 2*[n=0] + Fibonacci(n-2). (End)

A193737 Mirror of the triangle A193736.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 4, 3, 1, 3, 8, 8, 4, 1, 5, 15, 19, 13, 5, 1, 8, 28, 42, 36, 19, 6, 1, 13, 51, 89, 91, 60, 26, 7, 1, 21, 92, 182, 216, 170, 92, 34, 8, 1, 34, 164, 363, 489, 446, 288, 133, 43, 9, 1, 55, 290, 709, 1068, 1105, 826, 455, 184, 53, 10, 1, 89, 509, 1362, 2266, 2619, 2219, 1414, 682, 246, 64, 11, 1
Offset: 0

Views

Author

Clark Kimberling, Aug 04 2011

Keywords

Comments

This triangle is obtained by reversing the rows of the triangle A193736.

Examples

			First six rows:
  1;
  1,  1;
  1,  2,  1;
  2,  4,  3,  1;
  3,  8,  8,  4,  1;
  5, 15, 19, 13,  5,  1;
		

Crossrefs

Cf. A000007, A011782 (diagonal sums), A019590, A052542 (row sums).

Programs

  • Magma
    function T(n,k) // T = A193737
      if k lt 0 or n lt 0 then return 0;
      elif n lt 3 then return Binomial(n,k);
      else return T(n - 1, k) + T(n - 1, k - 1) + T(n - 2, k);
      end if;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 24 2023
    
  • Mathematica
    (* First program *)
    z=20;
    p[0, x_]:= 1;
    p[n_, x_]:= Fibonacci[n+1, x] /; n > 0
    q[n_, x_]:= (x + 1)^n;
    t[n_, k_]:= Coefficient[p[n, x], x^(n-k)];
    t[n_, n_]:= p[n, x] /. x -> 0;
    w[n_, x_]:= Sum[t[n, k]*q[n-k+1, x], {k,0,n}]; w[-1, x_] := 1;
    g[n_]:= CoefficientList[w[n, x], {x}]
    TableForm[Table[Reverse[g[n]], {n, -1, z}]]
    Flatten[Table[Reverse[g[n]], {n,-1,z}]] (* A193736 *)
    TableForm[Table[g[n], {n,-1,z}]]
    Flatten[Table[g[n], {n,-1,z}]]          (* A193737 *)
    (* Additional programs *)
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[1 + 1/(1 - x - x^2), 11]//Flatten  (* Peter Luschny, Feb 27 2021 *)
    T[n_, k_]:= T[n, k]= If[n<3, Binomial[n, k], T[n-1,k] + T[n-1,k-1] + T[n-2,k]];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 24 2023 *)
  • SageMath
    def T(n,k): # T = A193737
        if (n<3): return binomial(n,k)
        else: return T(n-1,k) +T(n-1,k-1) +T(n-2,k)
    flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 24 2023

Formula

Write w(n,k) for the triangle at A193736. This is then given by w(n,n-k).
T(0,0) = T(1,0) = T(1,1) = T(2,0) = 1; T(n,k) = 0 if k<0 or k>n; T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k). - Philippe Deléham, Feb 13 2020
From G. C. Greubel, Oct 24 2023: (Start)
T(n, 0) = Fibonacci(n) + [n=0] = A324969(n+1).
T(n, n-1) = n, for n >= 1.
T(n, n-2) = A034856(n-1), for n >= 2.
T(2*n, n) = A330793(n).
Sum_{k=0..n} T(n,k) = A052542(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A011782(n).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k,k) = A019590(n). (End)
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